DraftRoman · DraftRoman · Jun 8, 2026 · Jun 8, 2026 · Jun 8, 2026 · Jun 8, 2026
diff --git a/Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py b/Sprint-1/Python/calculate_sum_and_product/calculate_sum_and_product.py
@@ -12,20 +12,20 @@ def calculate_sum_and_product(input_numbers: List[int]) -> Dict[str, int]:
         "sum": 10, // 2 + 3 + 5
         "product": 30 // 2 * 3 * 5
     }
-    Time Complexity:
-    Space Complexity:
+    Time Complexity: O(n) because we are iterating through the list once now instead of twice before
+    Space Complexity: O(1) because we are using a constant amount of space
     Optimal time complexity:
+    We can calculate the sum and the product in a single pass through the list
     """
     # Edge case: empty list
     if not input_numbers:
         return {"sum": 0, "product": 1}
 
-    sum = 0
-    for current_number in input_numbers:
-        sum += current_number
-
+    total = 0     
     product = 1
+
     for current_number in input_numbers:
         product *= current_number
+        total += current_number
 
-    return {"sum": sum, "product": product}
+    return {"sum": total, "product": product}
diff --git a/Sprint-1/Python/find_common_items/find_common_items.py b/Sprint-1/Python/find_common_items/find_common_items.py
@@ -9,13 +9,26 @@ def find_common_items(
     """
     Find common items between two arrays.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity:O(n*m*k) because we are iterating through both lists and then checking if the item is already in the common items list
+    Space Complexity:O(n) because we are storing the common items in a new list
+    Optimal time complexity: O(n+m) because we have to iterate through both lists at least once to find the common items
     """
-    common_items: List[ItemType] = []
-    for i in first_sequence:
-        for j in second_sequence:
-            if i == j and i not in common_items:
-                common_items.append(i)
-    return common_items
+    # common_items: List[ItemType] = []
+    # for i in first_sequence:
+    #     for j in second_sequence:
+    #         if i == j and i not in common_items:
+    #             common_items.append(i)
+    # return common_items
+
+
+    first_set = set(first_sequence)
+    second_set = set(second_sequence)
+    return list(first_set & second_set)
+
+    """
+    Time Complexity now: O(n+m) because we are iterating through both list once
+    Space Complexity now: O(n+m) because we are storing the common items in a new list and we are also creating two sets which take up space
+    Optimal time complexity: O(n+m) because we have to iterate through both lists at least once to find the common items
+    """
+
+
diff --git a/Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py b/Sprint-1/Python/has_pair_with_sum/has_pair_with_sum.py
@@ -7,12 +7,22 @@ def has_pair_with_sum(numbers: List[Number], target_sum: Number) -> bool:
     """
     Find if there is a pair of numbers that sum to a target value.
 
-    Time Complexity:
-    Space Complexity:
-    Optimal time complexity:
+    Time Complexity: O(n^2) because we are iterating through the list twice to find the pair of numbers that sum to the target value
+    Space Complexity: O(1) because we are not using any extra space
+    Optimal time complexity: O(n) because we can use a hash set to store the numbers we have seen so far
     """
-    for i in range(len(numbers)):
-        for j in range(i + 1, len(numbers)):
-            if numbers[i] + numbers[j] == target_sum:
-                return True
+    # for i in range(len(numbers)):
+    #     for j in range(i + 1, len(numbers)):
+    #         if numbers[i] + numbers[j] == target_sum:
+    #             return True
+    # return False
+    for num in numbers:
+        complement = target_sum - num
+        if complement in numbers:
+            return True
     return False
+
+"""
+Time Comlexity now: O(n) because we are iterating once through the list to find the pair
+Scape Complexity now: O(n) because we aren't using any extra space, we ca have some list for saving the numbers, but we don't need it here
+"""
diff --git a/Sprint-1/Python/remove_duplicates/remove_duplicates.py b/Sprint-1/Python/remove_duplicates/remove_duplicates.py
@@ -4,22 +4,46 @@
 
 
 def remove_duplicates(values: Sequence[ItemType]) -> List[ItemType]:
+    # for value in values:
+    #     is_duplicate = False
+    #     for existing in unique_items:
+    #         if value == existing:
+    #             is_duplicate = True
+    #             break
+    #     if not is_duplicate:
+    #         unique_items.append(value)
+
+    # return unique_items
+
     """
     Remove duplicate values from a sequence, preserving the order of the first occurrence of each value.
 
-    Time complexity:
-    Space complexity:
-    Optimal time complexity:
+    Time complexity: O(n^2) we are iterating for each item in the list and then we're doing another iteration th check duplicates
+    Space complexity: O(n) creating a new list 
+    Optimal time complexity: O(n) we can use set to check  for duplicates 
     """
-    unique_items = []
+
+    unique_items: list[ItemType] = []
+    seen = set()
 
     for value in values:
-        is_duplicate = False
-        for existing in unique_items:
-            if value == existing:
-                is_duplicate = True
-                break
-        if not is_duplicate:
+        if value not in seen:
+            seen.add(value)
             unique_items.append(value)
-
     return unique_items
+
+
+
+
+
+
+    # unique_items = list(set(values))
+
+    # return unique_items
+"""
+
+Time: O(n)
+Space: O(n)
+
+"""
+